AP Calculus BC

Does the Sum of sqrt(n)/(sqrt(n)+1) Converge? No

Q: Does the sum of sqrt(n)/(sqrt(n)+1) converge?

No. The terms tend to 1, so the nth term test gives divergence.

Q: Why does the nth term test apply here?

Because \lim_{n \to \infty} a_{n} = 1, which is not 0. The test needs only that the limit fails to be zero.

Q: What if the series were 1/(sqrt(n)+1) instead?

Those terms do tend to 0, so the nth term test is silent. Limit comparison with \frac{1}{\sqrt{n}} then gives divergence.

The sum of sqrt n over sqrt n plus 1 diverges. Top and bottom both grow like sqrt n, so every term sits just below 1 rather than fading away. A term limit of 1 is exactly what the nth term test needs, and the running total climbs without bound.

\sum_{n=1}^{\infty}\frac{\sqrt{n}}{\sqrt{n}+1}

Diverges

Settled by the nth term test for divergence.

The terms settle at 1, not 0

Divide the numerator and the denominator by $\sqrt{n}$ to see what one term does for large $n$ .

\frac{\sqrt{n}}{\sqrt{n}+1} = \frac{1}{1 + \frac{1}{\sqrt{n}}}

The correction $\frac{1}{\sqrt{n}}$ dies out, so the fraction closes in on $1$ .

\lim_{n \to \infty}\frac{\sqrt{n}}{\sqrt{n}+1} = 1 \neq 0

A series whose terms do not tend to zero cannot converge. That is the nth term test, and it closes the question here in one line.

The root hides nothing

A square root in a term makes students expect shrinking pieces. What decides the limit is how the top compares with the bottom, not whether roots appear. Both grow like $\sqrt{n}$ , so their ratio settles at $1$ .

The verdict is already fixed by that limit. Watching the running total climb is an illustration, not a proof, and nothing about the totals is what an exam wants to see.

Compare with a term that really does shrink

The series with terms 1 over sqrt n plus 1 also diverges, but for a different reason. Those terms do tend to 0, so the nth term test says nothing and a limit comparison with 1 over sqrt n is the natural next step.

The mistakes students make

Every error here comes from misreading the size of the terms.

Reading $\sqrt{n}$ as a shrinking factor. It grows, and it grows at the same rate above and below the line.
Arguing that the denominator is larger than the numerator, so the terms must tend to $0$ . Staying below $1$ is not the same as approaching $0$ .
Skipping to the comparison tests. Direct comparison does settle it, since $a_{n} \ge \frac{1}{2}$ for every $n$ and $\sum \frac{1}{2}$ diverges, but the nth term limit is nonzero, so nothing else is needed.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the sum of sqrt(n)/(sqrt(n)+1) converge?

No. The terms tend to $1$ , so the nth term test gives divergence.

Why does the nth term test apply here?

Because $\lim_{n \to \infty} a_{n} = 1$ , which is not $0$ . The test needs only that the limit fails to be zero.

What if the series were 1/(sqrt(n)+1) instead?

Those terms do tend to $0$ , so the nth term test is silent. Limit comparison with $\frac{1}{\sqrt{n}}$ then gives divergence.